3.2625 \(\int \frac {x^{-1+3 n}}{(a+b x^n)^2} \, dx\)

Optimal. Leaf size=48 \[ -\frac {a^2}{b^3 n \left (a+b x^n\right )}-\frac {2 a \log \left (a+b x^n\right )}{b^3 n}+\frac {x^n}{b^2 n} \]

[Out]

x^n/b^2/n-a^2/b^3/n/(a+b*x^n)-2*a*ln(a+b*x^n)/b^3/n

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Rubi [A]  time = 0.03, antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {266, 43} \[ -\frac {a^2}{b^3 n \left (a+b x^n\right )}-\frac {2 a \log \left (a+b x^n\right )}{b^3 n}+\frac {x^n}{b^2 n} \]

Antiderivative was successfully verified.

[In]

Int[x^(-1 + 3*n)/(a + b*x^n)^2,x]

[Out]

x^n/(b^2*n) - a^2/(b^3*n*(a + b*x^n)) - (2*a*Log[a + b*x^n])/(b^3*n)

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

\begin {align*} \int \frac {x^{-1+3 n}}{\left (a+b x^n\right )^2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^2}{(a+b x)^2} \, dx,x,x^n\right )}{n}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {1}{b^2}+\frac {a^2}{b^2 (a+b x)^2}-\frac {2 a}{b^2 (a+b x)}\right ) \, dx,x,x^n\right )}{n}\\ &=\frac {x^n}{b^2 n}-\frac {a^2}{b^3 n \left (a+b x^n\right )}-\frac {2 a \log \left (a+b x^n\right )}{b^3 n}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 38, normalized size = 0.79 \[ \frac {-\frac {a^2}{a+b x^n}-2 a \log \left (a+b x^n\right )+b x^n}{b^3 n} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(-1 + 3*n)/(a + b*x^n)^2,x]

[Out]

(b*x^n - a^2/(a + b*x^n) - 2*a*Log[a + b*x^n])/(b^3*n)

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fricas [A]  time = 0.86, size = 59, normalized size = 1.23 \[ \frac {b^{2} x^{2 \, n} + a b x^{n} - a^{2} - 2 \, {\left (a b x^{n} + a^{2}\right )} \log \left (b x^{n} + a\right )}{b^{4} n x^{n} + a b^{3} n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1+3*n)/(a+b*x^n)^2,x, algorithm="fricas")

[Out]

(b^2*x^(2*n) + a*b*x^n - a^2 - 2*(a*b*x^n + a^2)*log(b*x^n + a))/(b^4*n*x^n + a*b^3*n)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3 \, n - 1}}{{\left (b x^{n} + a\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1+3*n)/(a+b*x^n)^2,x, algorithm="giac")

[Out]

integrate(x^(3*n - 1)/(b*x^n + a)^2, x)

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maple [A]  time = 0.03, size = 59, normalized size = 1.23 \[ -\frac {2 a \ln \left (b \,{\mathrm e}^{n \ln \relax (x )}+a \right )}{b^{3} n}+\frac {\frac {{\mathrm e}^{2 n \ln \relax (x )}}{b n}-\frac {2 a^{2}}{b^{3} n}}{b \,{\mathrm e}^{n \ln \relax (x )}+a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3*n-1)/(b*x^n+a)^2,x)

[Out]

(1/b/n*exp(n*ln(x))^2-2*a^2/b^3/n)/(b*exp(n*ln(x))+a)-2*a/b^3/n*ln(b*exp(n*ln(x))+a)

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maxima [A]  time = 0.51, size = 61, normalized size = 1.27 \[ \frac {b^{2} x^{2 \, n} + a b x^{n} - a^{2}}{b^{4} n x^{n} + a b^{3} n} - \frac {2 \, a \log \left (\frac {b x^{n} + a}{b}\right )}{b^{3} n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1+3*n)/(a+b*x^n)^2,x, algorithm="maxima")

[Out]

(b^2*x^(2*n) + a*b*x^n - a^2)/(b^4*n*x^n + a*b^3*n) - 2*a*log((b*x^n + a)/b)/(b^3*n)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {x^{3\,n-1}}{{\left (a+b\,x^n\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3*n - 1)/(a + b*x^n)^2,x)

[Out]

int(x^(3*n - 1)/(a + b*x^n)^2, x)

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sympy [A]  time = 157.50, size = 129, normalized size = 2.69 \[ \begin {cases} \frac {\log {\relax (x )}}{a^{2}} & \text {for}\: b = 0 \wedge n = 0 \\\frac {x^{3 n}}{3 a^{2} n} & \text {for}\: b = 0 \\\frac {\log {\relax (x )}}{\left (a + b\right )^{2}} & \text {for}\: n = 0 \\- \frac {2 a^{2} \log {\left (\frac {a}{b} + x^{n} \right )}}{a b^{3} n + b^{4} n x^{n}} - \frac {2 a^{2}}{a b^{3} n + b^{4} n x^{n}} - \frac {2 a b x^{n} \log {\left (\frac {a}{b} + x^{n} \right )}}{a b^{3} n + b^{4} n x^{n}} + \frac {b^{2} x^{2 n}}{a b^{3} n + b^{4} n x^{n}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(-1+3*n)/(a+b*x**n)**2,x)

[Out]

Piecewise((log(x)/a**2, Eq(b, 0) & Eq(n, 0)), (x**(3*n)/(3*a**2*n), Eq(b, 0)), (log(x)/(a + b)**2, Eq(n, 0)),
(-2*a**2*log(a/b + x**n)/(a*b**3*n + b**4*n*x**n) - 2*a**2/(a*b**3*n + b**4*n*x**n) - 2*a*b*x**n*log(a/b + x**
n)/(a*b**3*n + b**4*n*x**n) + b**2*x**(2*n)/(a*b**3*n + b**4*n*x**n), True))

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